Question

If x =root2+1/root 2-1 and y= root 2-1/root 2+1.Find the value of x^2+y^2+xy

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1} \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1}{ \sqrt{2} + 1} \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{2}) }^{2} + {(1)}^{2} + 2 \times \sqrt{2} \times 1}{ {( \sqrt{2} )}^{2} - {(1)}^{2} } \\ \\ = \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} \\ \\x = 3 + 2 \sqrt{2} \\ \\ y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1} \\ \\ using \: the \: identities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1}{ {( \sqrt{2}) }^{2} - {(1)}^{2} } \\ \\ = \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} \\ \\ y = 3 - 2 \sqrt{2} \\ \\ {x}^{2} + {y}^{2} + xy \\ \\ = {(3 + 2 \sqrt{2}) }^{2} + {(3 - 2 \sqrt{2}) }^{2} + (3 + 2 \sqrt{2} )(3 - 2 \sqrt{2} ) \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = ( {(3)}^{2} + {(2 \sqrt{2} }^{2} ) + 2 \times 3 \times 2 \sqrt{2} ) + ( {(3)}^{2} + {(2 \sqrt{2)} }^{2} - 2 \times 3 \times 2 \sqrt{2} ) + ({(3)}^{2} - {(2 \sqrt{2)} }^{2}) \\ \\ = (9 + 8 + 12 \sqrt{2} ) + (9 + 8 - 12 \sqrt{2} ) + (9 - 8) \\ \\ = (17 + 12 \sqrt{2} ) + (17 - 12 \sqrt{2} ) + 1 \\ \\ = 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2} + 1 \\ \\ = 17 + 17 + 1 + 12 \sqrt{2} - 12 \sqrt{2} \\ \\ ( + 12 \sqrt{2} \: and \: - 12 \sqrt{2} \: got \: cancel) \\ \\ = 17 + 17 + 1 \\ \\ = 35$


Hope this helps!!!!

@Mahak24

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